The freezing point of water is 0.00‚°C at 1 atmosphere. How many grams of chromium(III) nitrate (238.0 g/mol), must be dissolved in 255.0 grams of water to reduce the freezing point by 0.450°C ? Refer to the table for the necessary boiling or freezing point constant. Solvent Formula Kb (°C/m) Kf (°C/m) Water H2O 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHCl3 3.67 x Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02 x
The freezing point of water is 0.00°C at 1 atmosphere.ÂÂ
How many grams of chromium(III) nitrate (238.0 g/mol), must be dissolved in 255.0 grams of water to reduce the freezing point by 0.450°C ? Refer to the table for the necessary boiling or freezing point constant.
Solvent Formula Kb (°C/m) Kf (°C/m)
Water H2O 0.512 1.86
Ethanol CH3CH2OH 1.22 1.99
Chloroform CHCl3 3.67 x
Benzene C6H6 2.53 5.12
Diethyl ether CH3CH2OCH2CH3 2.02 x
_________g chromium(III) nitrate.
The Correct Answer and Explanation is :
To solve this problem, we need to use the concept of freezing point depression, which can be calculated using the formula:
[
\Delta T_f = i \cdot K_f \cdot m
]
Where:
- (\Delta T_f) is the freezing point depression (0.450°C in this case),
- (i) is the van ‘t Hoff factor (the number of particles into which a solute dissociates in solution),
- (K_f) is the freezing point depression constant for the solvent (water in this case, with (K_f = 1.86 \, \text{°C/m})),
- (m) is the molality of the solution.
Step 1: Calculate the molality of the solution
We can rearrange the formula to solve for molality:
[
m = \frac{\Delta T_f}{i \cdot K_f}
]
We know that the freezing point depression (\Delta T_f = 0.450 \, \text{°C}), the freezing point depression constant (K_f = 1.86 \, \text{°C/m}), and the solute is chromium(III) nitrate (Cr(NO₃)₃).
Chromium(III) nitrate dissociates into 4 ions (1 chromium ion (Cr³⁺) and 3 nitrate ions (NO₃⁻)), so the van ‘t Hoff factor (i = 4).
Now, we can calculate molality (m):
[
m = \frac{0.450 \, \text{°C}}{4 \cdot 1.86 \, \text{°C/m}} = \frac{0.450}{7.44} = 0.0605 \, \text{mol/kg}
]
Step 2: Use molality to find the number of moles of chromium(III) nitrate
Molality is defined as the number of moles of solute per kilogram of solvent. We are given that the mass of water is 255.0 grams, or 0.2550 kilograms. The number of moles of chromium(III) nitrate required is:
[
\text{moles of Cr(NO₃)₃} = m \cdot \text{mass of solvent in kg} = 0.0605 \, \text{mol/kg} \cdot 0.2550 \, \text{kg} = 0.0154 \, \text{mol}
]
Step 3: Convert moles of Cr(NO₃)₃ to grams
To find the mass of chromium(III) nitrate, we multiply the number of moles by the molar mass of chromium(III) nitrate, which is 238.0 g/mol:
[
\text{mass of Cr(NO₃)₃} = 0.0154 \, \text{mol} \cdot 238.0 \, \text{g/mol} = 3.67 \, \text{g}
]
Final Answer:
To reduce the freezing point of 255.0 grams of water by 0.450°C, you must dissolve 3.67 grams of chromium(III) nitrate.
Explanation:
In this problem, we used the freezing point depression equation, which allows us to calculate how much of a solute is needed to lower the freezing point of a solvent. The freezing point depression depends on the molality of the solution, which requires knowing the number of moles of solute and the mass of solvent. Since chromium(III) nitrate dissociates into 4 ions, we accounted for this factor (i = 4). Finally, by calculating the number of moles needed and converting this into grams using the molar mass of chromium(III) nitrate, we found the required mass to be 3.67 grams.